PHP编程：PHP实现的简单mock json脚本分享

<?php
/**
?* Mock Json for Javascript
?*
?* @author soulteary
?* @date 2014-06-15
?*/
?
?
/**
?* 哀求接口字段：字符集
?*/
define('charset','charset');
?
/**
?* 哀求接口字段：回调函数名
?*/
define('callback','callback');
?
/**
?* 哀求接口字段：跨域字段
?*/
define('crossDomain','cross-domain');
?
?
/**
?* 输出mock数据
?* 如果存在mock.json文件,则数据从mock.js中获取
?*
?* @return string
?*/
functionmockData()
{
??? if(file_exists('mock.json')){
??????? $data=json_decode(file_get_contents('mock.json'));
??? }else{
??????? $data=Array(
??????????? 'code'=>200,
??????????? 'desc'=>'Get the default data.',
??????????? 'login'=>true,
??????????? 'data'=>Array(
??????????????? 'name'=>'test api.'
??????????? )
??????? );
??? }
??? returnjson_encode($data);
}
?
?
/**
?* 输出字符集,允许结果为gbk、gb2312、utf-8
?* 如果非法或者未设置,输出utf-8
?*
?* @return string
?*/
functioncharset()
{
??? $ret='utf-8';
??? if(empty($_REQUEST[charset])){
??????? return$ret;
??? }else{
??????? $charset=strtolower($_REQUEST[charset]);
??????? if(in_array($charset,array('gbk','gb2312'),true)){
??????????? return$charset;
??????? }else{
??????????? return$ret;
??????? }
??? }
}
?
?
/**
?* 拼装json数据
?*
?* @return string
?*/
functionjsonGenerator()
{
??? if(!empty($_REQUEST[callback])){
??????? header('Content-Type: application/javascript; charset='.charset());
??????? return$_REQUEST[callback]."(".mockData().");";
??? }else{
??????? if(!empty($_REQUEST[crossDomain])){
??????????? header("Access-Control-Allow-Origin: *");
??????? };
??????? header('Content-type: application/json; charset='.charset());
??????? returnmockData();
??? }
}
?
/**
?* 输出结果
?*/
die(jsonGenerator());

{
??? "data":1,
??? "w":"测试"
}